Maybe l = filter(a, lambda v: v == a[-1]) sorted(a[:-len(l)]) + l
? On Fri, Sep 28, 2012 at 4:51 PM, dave <davidrey...@gmail.com> wrote: > more clearer, this is a more realistic use case: > > ['awefawef', 'awefawfsf', 'awefsdf', 'zzzzzzzzzzzzzz', 'zzzzzzzzzzzzzz', > 'zzzzzzzzzzzzzz'] > > and the quantity of ''zzzzzzzzzzzzzz'' would be dynamic. > > On Friday, September 28, 2012 4:46:15 PM UTC-7, Ian wrote: > > > > > > a = ['a', 'b', x] > > > > > > > > > > b = sorted(a) > > > > > > > > > > What does x need to be to always be last on an ascending sort no > matter what 'a' and 'b' are.... within reason... I am expecting 'a' and 'b' > will be not longer than 10 char's long.... I tried making x = > 'zzzzzzzzzzzzzzzz' and believe it or not, this appears FIRST on the sort!!! > > > > > > > > It appears last when I run the code. > > > > > > > > To answer your question, though, if you want to force x to be last, > > > > then I suggest removing it from the list and then appending it to the > > > > end. > -- > http://mail.python.org/mailman/listinfo/python-list >
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